📂Number Theory

Identity for Dirichlet Products

Definition ¹

A arithmetic function defined as follows $I$ is called the identity function. $$I(n) := \left[ {{ 1 } \over { n }} \right]$$

• [1] Identity series: This is the unit function $u$. In other words, $$\sum_{d \mid n}I(d) = u(n) = 1$$
• [2] Completely multiplicative: For all $n , m \in \mathbb{N}$, $I (mn) = I(m) I(n)$
• [a] Identity element for convolution: For all arithmetic functions $f$, $$I \ast\ f = f \ast\ I = f$$

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• $\left[ x \right] = \lceil x \rceil$ is called the floor function and represents the largest integer less than or equal to $x$.

Explanation

$$\begin{matrix} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ I(n) & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \sum_{d \mid n} I(d) & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{matrix}$$ In most of mathematics, the name identity function is given to the function whose elements of the domain map to themselves as in $i(x) = x$, but at least in analytic number theory, it’s called the norm $N (n) = n$. As seen in $I$, it can be called an identity because it always exists as the identity element for convolution $\ast$.

Proof

[1]

$\displaystyle I(n) = \left[ {{ 1 } \over { n }} \right] = \begin{cases} 1 & , n=1 \\ 0 &, n>1 \end{cases}$ is true. Therefore, $$\sum_{d \mid n}I(d) = 1 + 0 + \cdots = 1$$

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[2]

• Case 1. $m = n = 1$ $$I ( mn ) = I(1) = 1 = 1 \cdot 1 = I(1) I(1) = I(m) I(n)$$
• Case 2. $m = 1 \land n > 1$ $$I(mn) = I (n) = 1 \cdot I (n) = I(m) I(n)$$
• Case 3. $m > 1 \land n = 1$ $$I(mn) = I (m) = I(m) \cdot 1 = I(m) I(n)$$
• Case 4. $m > 1 \land n > 1$ $$I(mn) = 0 = 0 \cdot 0 = I(m) I(n)$$

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[a]

Since $d$ is a divisor of $n$, in the case of $d \ne n$, $\displaystyle \left[ {{ d } \over { n }} \right] = 0$ and $$(f \ast\ I)(n) = \sum_{d \mid n} f(d) I \left( {{ n } \over { d }} \right) = \sum_{d \mid n} f(d) \left[ {{ d } \over { n }} \right] = f(n)$$ According to the commutative law of convolution of arithmetic functions, $f \ast\ I = I \ast\ f = f$

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